Which interval contains a local minimum for the grafted function? [-4, -2.5] [-2, -1] [1, 2] [2.5, 4]
1 answer:
Answer:
Last option [2.5, 4]
Step-by-step explanation:
The local minima of a function are the points where the slope of the curve is zero and the function reaches a minimum value.
If, on the contrary, the function reaches a maximum value, then that point is a local maximum.
Observe, for example, the point (3, -4). Note that a line tangent to that point would be horizontal, that is, of slope m = 0. This point is a minimum of the function, because there are no values x to the right x = 3 or to the left of x = 3 for which
Now we must search among the options that interval contains at this point.
The intervals [-4, -2.5] [-2, -1] do not contain any local minimum
The intervalor [1, 2] contains a local maximum.
The only interval that contains a local minimum is [2.5, 4], which contains the point (3, -4)
You might be interested in
Answer:
Null hypothesis is
Alternative hypothesis is
Test Statistics z = 2.65
CONCLUSION:
Since test statistics is greater than critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.
P- value = 0.004025
Step-by-step explanation:
Given that:
Mean = 960 hours
Sample size n = 36
Mean population 937
Standard deviation = 52
Given that the mean time between failures is 937 hours. The objective is to determine if the mean time between failures is greater than 937 hours
Null hypothesis is
Alternative hypothesis is
Degree of freedom = n-1
Degree of freedom = 36-1
Degree of freedom = 35
The level of significance ∝ = 0.01
SInce the degree of freedom is 35 and the level of significance ∝ = 0.01;
from t-table t(0.99,35), the critical value = 2.438
The test statistics is :
Z = 2.65
The decision rule is to reject null hypothesis if test statistics is greater than critical value.
CONCLUSION:
Since test statistics is greater than critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.
The P-value can be calculated as follows:
find P(z < - 2.65) from normal distribution tables
= 1 - P (z ≤ 2.65)
= 1 - 0.995975 (using the Excel Function: =NORMDIST(z))
= 0.004025